Question

The correct simplification of the quadratic formula used to solve 8x 2 - 2x = 1?

Answer 1

Answer:

$x_{1}=\frac{1}{2}\\x_{2}=-\frac{1}{4}$

Step-by-step explanation:

Hello

For ax2 + bx + c = 0,the values of x which are the solutions of the equation are given by

$x=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

± means  there are two solution for x,  X1 and X2

$x_{1} =\frac{-b+\sqrt{b^{2}-4ac } }{2a} \\x_{2} =\frac{-b-\sqrt{b^{2}-4ac } }{2a}$

Step 1

convert 8x2-2x=1 into the form ax2 + bx + c = 0 ( right side equal to cero)

subtract 1 in each side

$8x^{2} -2x=1\\8x^{2} -2x-1=1-1\\8x^{2} -2x-1=0$

Step 2

replace in the equation

Let

a=8

b=-2

c=-1

Hence

$x_{1} =\frac{-b+\sqrt{b^{2}-4ac }}{2a}\\ x_{1}=\frac{-(-2)+\sqrt{(-2)^{2}-4(8)(-1) } }{2*8}\\x_{1} =\frac{+2+\sqrt{4+32 }}{16}\\x_{1} =\frac{+2+\sqrt{36 } }{16}\\x_{1} =\frac{2+\sqrt{36 }}{16}\\x_{1} =\frac{2+6}{16} \\x_{1} =\frac{8}{16} \\x_{1}=-\frac{1}{2}$

Now, X2

$x_{2} =\frac{-b-\sqrt{b^{2}-4ac }}{2a}\\ x_{2}=\frac{-(-2)-\sqrt{(-2)^{2}-4(8)(-1) } }{2*8}\\x_{1} =\frac{+2-\sqrt{4+32 }}{16}\\x_{2} =\frac{+2-\sqrt{36 } }{16}\\x_{2} =\frac{2-\sqrt{36 }}{16}\\x_{2} =\frac{2-6}{16} \\x_{2} =\frac{-4}{16} \\x_{2}=-\frac{1}{4}$

Have a great day

Answer 2

(2x−1)(4x+1)..................hope this helps